Let P(a secθ,b tanθ) and Q(a secϕ,b tanϕ), where θ+ϕ=π2, be two points on the hyperbola x2a2−y2b2=1.
If (h, k) is the point of the intersection of the normals at P and Q, then k is equal to
−(a2+b2b)
Firstly, we obtain the slope of normal to x2a2−y2b2=1 at (a secθ,b tanθ)
On differentiating w.r.t.x, we get
2xa2−2yb2×dydx=0⇒dydx=b2a2xy
Slope for normal at the point (a secθ,b tanθ) will be −a2b tanθb2a secθ=−absinθ
∴ Equation of normal at (a secθ,b tanθ) is
y−b tanθ=−absinθ(x−a secθ)
⇒(a sinθ)x+by=(a2+b2)tanθ
⇒ax+b y cosecθ=(a2+b2)secθ ... (i)
Similarly, equation of normal to x2a2−y2b2=1 at (a secϕ,b tanϕ) is ax+b y cosec ϕ=(a2+b2)secϕ ... (ii)
On subtracting Eq. (ii) from Eq. (i), we get
b(cosecθ−cosecϕ)y=(a2+b2)(secθ−secϕ)
⇒y=a2+b2b.secθ−secϕcosecθ−cosecϕ
But secθ−secϕcosecθ−cosecϕ=secθ−sec(π2−θ)cosecθ−cosec(π2−θ)
[∵ϕ+θ=π2]
=secθ−cosecθcosecθ−secθ=−1
Thus, y=−(a2+b2b),i.e.k=−(a2+b2b)